Question

An incompressible fluid flows steadily through a cylindrical pipe which has radius $2R$ at point A and radius $R$ at point B farther along the flow direction. If the velocity at point $A$ is $v$, its velocity at point B is

• A. $2v$
• B. $v$
• C. $\frac{v}{2}$
• D. $4v$

Answer 1

The correct option is D $4v$

We know, for the fluid flowing through the nonuniform pipe the velocity of a fluid is inversely proportional to the area of cross-section.
Hence, according to problem given, if $v_1,v_2$ are the velocities at A and B and $a_1,a_2$ are the area of cross-sections at A and B, then
$\frac{v_2}{v_1}=\frac{a_1}{a_2}$
Here, an in-compressible fluid flows steadily through a cylindrical pipe which has radius $2R$ at point A and radius $R$ at point B farther along the flow direction, hence
$v_2=\frac{a_1{v}_1}{a_2}=\frac{\pi \left(2R{)}^2\right(v)}{\pi R^2}$
$\therefore v_2=4v$